In Fig 34-58 a pinecone is at distance p₁ =1.0min front of a lens of focal length f₁=0.50m; a flat mirror is at distance d=2.0mbehind the lens. Light from the pinecone passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the pinecone. What are (a) the distance between the lens and that image and (b) the overall lateral magnification of the pinecone? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted relative to the pinecone or not inverted?

Object distance p₁=1.0m

Focal length of lens f₁=0.50m

The distance of mirror from lens is, d=2.0

We use the mirror equation to find the image distance. As the image ofthelens istheacting object for the mirror, we can usethelens equation again to get the final image position. We need to ensure that we consider the mirror-lens separation distance as well.

Formula:

$$\begin{gathered} m=-\frac{i}{p} \\ \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \end{gathered}$$

(a)

The mirror equation relates an object distance p₁, mirror’s focal length f₁ and the image distance i₁ due to the first mirror:

$$\begin{gathered} \frac{1}{p_1}+\frac{1}{i_1}=\frac{1}{f_1} \\ \frac{1}{1}+\frac{1}{i_1}=\frac{1}{0.50} \\ \frac{1}{i_1}=\frac{1}{0.50}-\frac{1}{1.0} \\ {i}_1=1.0m \end{gathered}$$

Hence the image formed by the converging lens is located at distance 1.0m to the right of the lens.

And this image is at a distance 2m-1m=1m to the left of mirror. This image is real. The image formed by the mirror for this real image is then at 1m to the right of the mirror. This other image is then

p' =2m+1m

=3m

to the right side of the lens.

This image then acts as an object for the lens and results in another image formed by the lens, located at distance(i').

$$\begin{gathered} \frac{1}{p\text{'}}+\frac{1}{i\text{'}}=\frac{1}{f_1} \\ \frac{1}{3m}+\frac{1}{i\text{'}}=\frac{1}{0.50m} \\ i\text{'}=0.60m \end{gathered}$$

$$\begin{gathered} \frac{1}{p\text{'}}+\frac{1}{i\text{'}}=\frac{1}{f_1} \\ \frac{1}{3m}+\frac{1}{i\text{'}}=\frac{1}{0.50m} \\ i\text{'}=0.60m \end{gathered}$$

This image is formed to the left of the lens and is at a distance 2m+0.60m=2.60m at the left from the mirror.

(b)

The magnification ofthereal first image is

$$\begin{gathered} m_1=-\frac{i_1}{p_1} \\ =-\frac{1m}{1m} \\ =-1 \end{gathered}$$

The magnification of image 2 is

$$\begin{gathered} m_2=-\frac{i\text{'}}{p\text{'}} \\ =-\frac{0.60m}{3m} \\ =-0.20 \end{gathered}$$

The net lateral magnification m oftheobject is

m=m₁x m₂

$\begin{array}{l}m=\frac{i_1}{p_1}\times \frac{i^i}{p^{\text{'}}}\\ =-1\times -0.20\\ =+0.20\end{array}$

The overall lateral magnification of pinecone is +0.20.

(c)

Since i' is positive, the final image formed is real.

(d)

The image formed is to the left of the lens.

(e)

The lateral magnification m oftheobjectis positive; hence the image has the same orientation as the object, that is, the image is not inverted.